Answer to Question #153496 in Quantitative Methods for usman

Question

Given f(0) = – 18, f(1) = 0, f(3) = 0, f(5) = – 248, f(6) = 0, f(9) = 13104; find f(x).

Answer 1

$$\begin{matrix} x_0=0 & y_0=-18\ \ \\ x_1=1 & y_1=0\ \ \ \ \ \ \ \\ x_2=3 & y_2=0\ \ \ \ \ \ \ \\ x_3=5 & y_3=-248\\ x_4=6 & y_4=0\ \ \ \ \ \ \ \\ x_5=9 & y_5=13104\\ \end{matrix}$$

The interpolating polynomial is:

$$L(x)=\dfrac{(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)(x_0-x_4)(x_0-x_5)}\times y_0$$

$$+\dfrac{(x-x_0)(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_1-x_5)}\times y_1$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_3)(x-x_4)(x-x_5)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)(x_2-x_4)(x_2-x_5)}\times y_2$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_2)(x-x_4)(x-x_5)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)(x_3-x_4)(x_3-x_5)}\times y_3$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_5)}{(x_4-x_0)(x_4-x_1)(x_4-x_2)(x_4-x_3)(x_4-x_5)}\times y_4$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_4)}{(x_5-x_0)(x_5-x_1)(x_5-x_2)(x_5-x_3)(x_5-x_4)}\times y_5$$

$$L(x)=\dfrac{(x-1)(x-3)(x-5)(x-6)(x-9)}{(0-1)(0-3)(0-5)(0-6)(0-9)}\times (-18)$$

$$+0$$

$$+0$$

$$+\dfrac{(x-0)(x-1)(x-3)(x-6)(x-9)}{(5-0)(5-1)(5-3)(5-6)(5-9)}\times (-248)$$

$$+0$$

$$+\dfrac{(x-0)(x-1)(x-3)(x-5)(x-6)}{(9-0)(9-1)(9-3)(9-5)(9-6)}\times 13104$$

$$L(x)=(x^2+x+1)(x-1)(x-3)(x-6)$$